14. Compute the limit by using the definition of the derivative. $$ \lim _{ heta ightarrow \frac{\pi}{6}} \frac{\sin heta-\frac{1}{2}}{ heta-\frac{\pi}{6}} $$

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14. Compute the limit by using the definition of the derivative. $$ \lim _{	heta ightarrow \frac{\pi}{6}} \frac{\sin 	heta-\frac{1}{2}}{	heta-\frac{\pi}{6}} $$

UpStudy AI · Accepted Answer

We recognize that the limit is of the form of the definition of the derivative for the function $$ f(\theta)=\sin \theta $$ at the point $$ \theta=\frac{\pi}{6} $$.

1. The derivative of $$ f $$ at $$ \theta=\frac{\pi}{6} $$ is given by
   $$
   f'\left(\frac{\pi}{6}\right) = \lim_{\theta \rightarrow \frac{\pi}{6}} \frac{f(\theta)-f\left(\frac{\pi}{6}\right)}{\theta-\frac{\pi}{6}}.
   $$

2. Since $$ f(\theta)=\sin\theta $$ and $$ \sin\left(\frac{\pi}{6}\right)=\frac{1}{2} $$, the limit becomes
   $$
   \lim_{\theta \rightarrow \frac{\pi}{6}} \frac{\sin \theta-\frac{1}{2}}{\theta-\frac{\pi}{6}} = f'\left(\frac{\pi}{6}\right).
   $$

3. The derivative of $$ \sin \theta $$ is $$ \cos \theta $$. Thus,
   $$
   f'\left(\frac{\pi}{6}\right) = \cos \left(\frac{\pi}{6}\right).
   $$

4. Evaluating $$ \cos\left(\frac{\pi}{6}\right) $$ we have
   $$
   \cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}.
   $$

Therefore, the limit is
$$
\frac{\sqrt{3}}{2}.
$$
The limit is $$ \frac{\sqrt{3}}{2} $$.

Compute the limit by using the definition of the derivative.

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